 Solution

Let $\psi$ have energy $E_n$ and put $\gamma =2m\hslash \omega$. The lowering norm formula gives, inductively,In particular $(a_{-}{)}^n\psi$ is a nonzero ground state. By the assumed nondegeneracy, it equals $c{\psi }_0$ for some $c\ne 0$.

It remains to show that raising this vector recovers $\psi$, rather than merely another state at the same energy. On an energy eigenstate at $E_n$, factorization gives $a_{+}{a}_{-}\psi =\gamma n\psi$. Iterating the factorization down the ladder yieldsApplying $(a_{+}{)}^n$ to $(a_{-}{)}^n\psi =c{\psi }_0$ therefore provesFor $n=0$ this is the original ground-state assumption. This proves nondegeneracy propagated by an oscillator ladder: every energy eigenspace is one-dimensional if the ground state is nondegenerate.